Kirchhoff’s first rule – problems and solutions
1. Based on figure as shown below, if I1 = 5A, I2 = 3 A, I3 = 6 A, then I
Solution :
Kirchhoff’s first rule states that the sums of all currents entering the junction must equal the sum of all currents leaving the junction.
I1 and I2 are entering the junction, whereas I3 and I4 are leaving.
Apply Kirchhoff’s first rule :
I1 + I2 = I3 + I4
5 + 3 = 6 + I4
8 = 6 + I4
I4 = 8 – 6
I4 = 2 Ampere
2. If I1 = I3 then I4 =……..
Known :
I1 = I3 = 2 Ampere
I2 = 3 Ampere
I5 = 4 Ampere
Wanted: I4
Solution :
I1 and I4 are entering the junction, whereas I2, I3, and I5 are leaving.
Apply Kirchhoff’s first rule :
I1 + I4 = I2 + I3 + I5
2 + I4 = 3 + 2 + 4
2 + I4 = 9
I4 = 9 – 2
I4 = 7 Ampere
3. I1 = …….
Solution :
I am entering the junction, whereas I1, I2, and I3 are leaving.
Apply Kirchhoff’s first rule :
I = I1 + I2 + I3
7 A = I1 + 1 A + 4 A
7 A = I1 + 5 A
7 A – 5 A = I1
I1 = 2 Ampere
4. I2 = ………
Wanted: I2
Solution :
I1 is entering the junction, whereas I2, I3, and I4 are leaving.
Apply Kirchhoff’s first rule :
I1 = I2 + I3 + I4
6 = I2 + 3 + 1
6 = I2 + 4
I2 = 6 – 4
I2 = 2 Ampere
5. If I2 = 1/4 I1 , then I3 = ………
Known :
I1 = 600 milliAmpere
Wanted: I3
Solution :
I2 = 1/4 I1 = 1/4 (600 mA) = 150 mA
I1 is entering the junction, whereas I2 and I3 are leaving.
Apply Kirchhoff’s first rule :
I1 = I2 + I3
600 mA = 150 mA + I3
I3 = 600 mA – 150 mA
I3 = 450 mA
6. Based on figure below, find I!
Solution :
Apply Kirchhoff‘s first rule :
3 + 5 + 2 = 7 + I
10 = 7 + I
10 – 7 = I
I = 3 Ampere
7. Based on figure below, if I1 = 4 A, I2 = 3 A, I4 = 7 A, I5 = 4 A, determine I3.
Solution :
Electric current entering the junction = I1 + I3 = 4 A + I3
Electric current leaving the junction = I2 + I4 + I5 = 3 A + 7 A + 4 A = 14 A
14 A = 4 A + I3
I3 = 14 A – 4 A
I3 = 10 A